A projectile is shot straight up from the earth's surface at a speed of 1.50×104 .
What are the speed units? You have not asked a question.
af = mc#2
To determine the step-by-step details of the projectile's motion, we need some additional information. Can you provide the unit of speed (m/s or km/h) and specify if there are any other parameters, such as the direction of the acceleration or the angle at which it is shot?
To calculate the time it takes for the projectile to reach the maximum height and fall back to the ground, we need to use the equations of motion.
The initial velocity of the projectile, u, is 1.50×10^4 m/s. Since it is shot straight up, the final velocity at the maximum height is 0 m/s.
Using the equation:
v = u + at
where v is the final velocity, u is the initial velocity, a is the acceleration, and t is the time, we can determine the time it takes for the projectile to reach the maximum height.
Since the final velocity at the highest point is 0 m/s, we can rewrite the equation as:
0 = 1.50×10^4 m/s - 9.8 m/s^2 * t
Solving for t:
1.50×10^4 m/s = 9.8 m/s^2 * t
t = (1.50×10^4 m/s) / 9.8 m/s^2
Now we can calculate the time it takes for the projectile to reach the maximum height:
t = 1530.61 s (rounded to two decimal places)
This is the time it takes for the projectile to reach its maximum height. To calculate the total time it takes for the projectile to fall back to the ground, we need to double this time:
Total time = 2 * 1530.61 s = 3061.22 s (rounded to two decimal places)
Therefore, the projectile takes approximately 3061.22 seconds to reach the maximum height and fall back to the ground.